Let us show (REF ). By the Law of Large Numbers, \(\hat{s}(f) \rightarrow s(f)\)  as \(N \rightarrow \infty \)  in probability. Apply Slutsky's theorem: [1]}. From (REF ) and (REF ), we have convergence in law:
\(\begin{split}&N^{1/2}\Bigl [\hat{\rho }_k(f_i, f_j),\, i, j = 1, \ldots , m,\, k = 1, \ldots , K\Bigr ] \\ \Rightarrow \Bigl [\eta _{ijk} &:= \frac{\xi _{ijk}}{s(f_i)s(f_j)},\, i, j = 1, \ldots , m,\, k = 1, \ldots , K\Bigr ],\\ [\xi _{ijk},\, & i, j = 1, \ldots , m] \sim \mathcal {N}_{m^2}(\mathbf {0}, \mathbf {Q}),\quad k = 1, \ldots , K.\end{split}\) 

From (REF ), we get: \([\eta _{ijk}] \sim \mathcal {N}_{m^2}(\mathbf {0}, \mathbf {C})\)  i.i.d. for \(k = 1, \ldots , K\) , with
\((\mathbf {C})_{ij, i^{\prime }j^{\prime }} &:= \frac{\mathbf {Q}_{ij, i^{\prime }j^{\prime }}}{s(f_i)s(f_j)s(f_{i^{\prime }})s(f_{j^{\prime }})} = \frac{Q(f_i, f_{i^{\prime }})Q(f_{j}, f_{j^{\prime }})}{s(f_i)s(f_j)s(f_{i^{\prime }})s(f_{j^{\prime }})} \\ & = C(f_i, f_{i^{\prime }})C(f_j, f_{j^{\prime }}).\) 

In the last line, we used (REF ). This completes the proof of (REF ).
